Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{y - 5}{2y - 6} \times \dfrac{y^2 - 10y + 21}{3y - 15} $
Answer: First factor the quadratic. $r = \dfrac{y - 5}{2y - 6} \times \dfrac{(y - 3)(y - 7)}{3y - 15} $ Then factor out any other terms. $r = \dfrac{y - 5}{2(y - 3)} \times \dfrac{(y - 3)(y - 7)}{3(y - 5)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (y - 5) \times (y - 3)(y - 7) } { 2(y - 3) \times 3(y - 5) } $ $r = \dfrac{ (y - 5)(y - 3)(y - 7)}{ 6(y - 3)(y - 5)} $ Notice that $(y - 5)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ \cancel{(y - 5)}(y - 3)(y - 7)}{ 6\cancel{(y - 3)}(y - 5)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $r = \dfrac{ \cancel{(y - 5)}\cancel{(y - 3)}(y - 7)}{ 6\cancel{(y - 3)}\cancel{(y - 5)}} $ We are dividing by $y - 5$ , so $y - 5 \neq 0$ Therefore, $y \neq 5$ $r = \dfrac{y - 7}{6} ; \space y \neq 3 ; \space y \neq 5 $